document.write( "Question 933215: Please can someone help me I have been trying for hours!\r
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document.write( "A college entrance exam has a mean of 1400 and standard deviation of 200. A college will only accept the top 2.5% or higher. What is the lowest score they need on the exam? \r
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document.write( "Can someone please help and show the formula! I would appreciate it so much! \n" );
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Algebra.Com's Answer #566680 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! mean of 1400 and a standard deviation of 200. \n" ); document.write( " A college will only take applicants in the top 2.5% or higher. \n" ); document.write( " What is the lowest score they would need on the entrance exam \n" ); document.write( " 200z + 1400 = x \n" ); document.write( "z = invNorm(.975). \n" ); document.write( " invNorm(.975) = 1.96 can be found from a table or Using calculator 0r Excel Function \n" ); document.write( "..... \n" ); document.write( " 200(invNorm(.975) + 1400 = 200(1.96) + 1400 = 1792, lowest score \n" ); document.write( "..... \n" ); document.write( "Note: 1 - .025 = .975 (z-value found...97.5% of the Area Under NOrmal curve to the left of that value. \n" ); document.write( "...... \n" ); document.write( "For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted. \n" ); document.write( "97.5% of the Area under the standard normal curve is to the left of z = 1.96 \n" ); document.write( "Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right \n" ); document.write( " \n" ); document.write( "stattrek.com a great reference for statistics \n" ); document.write( " \n" ); document.write( " |